A 4-digit number contains at least one even digit. How many such numbers are there? Please help me. I will mark brainliest

Given:

A 4-digit number contains at least one even digit.

To find:

How many such numbers are there?

Solution:

Digits are 0,1,2,3,4,5,6,7,8,9.

Number of odd digits = 5

Number of even digits = 5

4-digit numbers are from 1000 to 9999.

Total number of 4-digit numbers = 9000

Possible ways to get odd digits on all the 4 places is

It means 625 numbers are their in which all 4 digits are odd.

To find total 4-digit number that contains at least one even digit, subtract the numbers which contains only odd numbers from the total 4-digit numbers.

Therefore, total 4-digit number that contains at least one even digit is 8375.

for the points

step-by-step explanation:

x² + 3x + 7 = 0       [ax² + bx + c]

a = 1

b = 3

c = 7

quadratic formula:

    (+/- is this ±)

plug in the numbers into the variables

your answer is a